An effort of 50 kgf is applied at the end of the lever of the second order, which supports a load of 750 kgf, such that the load is at a distance of 0.1 m from the hinge. Find the length of the lever.
[Assume lever is weightless]

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Solution

Step: 1
Given
AB = Load arm ( Distance of load from hinge) = 0.1 m
let AE = Effort arm or length of lever = y
Load = 750 kgf
Effort = 50 kgf
Step - 2
Diagram -
Step - 3
Formula used :
Effort x Effort arm = Load x Load arm
50 x y = 750 x 0.1
y = 75 / 50
y = 1.5 m
Hence length of lever = 1.5 m

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An effort of 50 kgf is applied at the end of the lever of the second order, which supports a load of 750 kgf, such that the load is at a distance of 0.1 m from the hinge. Find the length of the lever. [Assume lever is weightless]

Step: 1 Given AB = Load arm ( Distance of load from hinge) = 0.1 m let AE = Effort arm or length of lever = y Load = 750 kgf Effort = 50 kgfStep - 2 Diagram - Step - 3Formula used : Effort x Effort arm = Load x Load arm 50 x y = 750 x 0.1y = 75 / 50 y = 1.5 m Hence length of lever = 1.5 m

An effort of 50 kgf is applied at the end of the lever of the second order, which supports a load of 750 kgf, such that the load is at a distance of 0.1 m from the hinge. Find the length of the lever.
[Assume lever is weightless]

Step: 1
Given
AB = Load arm ( Distance of load from hinge) = 0.1 m
let AE = Effort arm or length of lever = y
Load = 750 kgf
Effort = 50 kgf
Step - 2
Diagram -
Step - 3
Formula used :
Effort x Effort arm = Load x Load arm
50 x y = 750 x 0.1
y = 75 / 50
y = 1.5 m
Hence length of lever = 1.5 m